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31st January 2013, 01:10 AM  #1 
diyAudio Member
Join Date: Jan 2013

amplifier output impedance measurement
I am attempting to understand more about how to measure the output impedance of an unknown amp, say the circlotron I have built. I know about the formula involving mu and all that, but I'm intrested in a more direct approach, as the circlotron's unusual output topology, the formula might not be accurate....
I posted earlier with my idea of an AC ammeter in series with a variable load resistor, and was informed by a number of fellow bloggers that the correct approach is an AC voltmeter across a variable resistor...both ideas were tried and failed. With the ammeter in series mode, current simply went up as resistance went down, as expected, however when applying the output to an AC voltmeter across a variable resistor, the voltage simply went up, with no peak, till open circut, produced the highest. I'm using white noise, from an FM tuner as signal source. With a music input, results look good, I'm getting 1012 watts of clean output into an 8 ohm speaker. Early on, I was using a 2000 ohm 10 watt universal output transformer, but as I've gotten more of the bugs worked out, direct to speaker is now louder than with the transformer, which is a little suprising. Louder than that and the driver stages distorts. I'm in the process of redesigning the driver stage using the Grommes model 260A as a reference. I'm using four 6550's (borrowed from my mac), but I have space for six, so two more are on order. I plan on rewiring my old school, bose 901 speakers, in series to get 64 ohms impedance. Could I be getting into trouble with such an arrangement with an uneven speaker impedance? Has anyone out there tried this arrangement of circlotron driving 64 ohm boses? I read that getting enough drive voltage to the output stage would be a problem, which I am finding even with nontriode outputs. Last edited by midmoe; 31st January 2013 at 01:21 AM. Reason: misspelled "speaker" 
31st January 2013, 02:12 AM  #2 
diyAudio Member
Join Date: Jun 2011

You measure the voltage across a resistor, call this V1 R1, then you measure the voltage across another smaller resistor (without changing the settings). Call this V2 R2. Then you calculate the output impedance:
Z=(V1V2)/((V2/R2)(V1/R1)) Last edited by counter culture; 31st January 2013 at 02:18 AM. 
31st January 2013, 02:39 AM  #3 
diyAudio Member
Join Date: Aug 2009
Location: College Station, TX

This was discussed, and, I thought, explained in the previous thread. The idea that I *think* you have in mind is that you want to adjust the variable load resistor until "something" is maximised. The problem is that what you would need to maximise is not the voltage across the resistor, nor is it the current flowing through the resistor. As you have now discovered, the unsurprising outcome if you try to maximise the current is that the resistance should be zero, whilst if you want to maximise the voltage the resistance should be infinity.
If you *really* want to try to measure the output impedance using this type of method, you must instead adjust the resistance so as to maximise the *power*. In other words, you would need to measure the current *and* the voltage, and then multiply them together to get the power. Then, adjust the resistor, remeasure the current and the voltage and multiply them together to get the new result for the power. If it is bigger than the previous value, then keep adjusting the resistor in the same direction, and try again. If it is smaller than the previous power, adjust the resistor in the opposite direction and try again. Keep on going like this, adjusting, remeasuring, recalculating, over and over again, until you home in on the resistance setting that maximises the power. A very clumsy and tedious process, and not at all to be recommended!!!! Instead, the easy way to do it is as follows. Since in your case you are seemingly expecting an output impedance in the ballpark of 64 ohms, then do the following. Use a 64 ohm dummy resistive load in place of the speaker. Feed in a sinewave signal to the amplifier, and measure the AC output signal voltage on the 64 ohm resistive load. Then, leaving the inut signal unaltered, add an additional resistor in parallel with the 64 ohm load, and remeasure the output voltage across the load. With these measurements, you can calculate the output impedance, as follows: To allow for more general cases, let's say the original resistive load you choose is R1 ohms. And that for the second measurement you connect an additional resistance R2 ohms in parallel with this. And suppose the output voltage you measure with the load R1 by itself is V1, and that the voltage after adding R2 in parallel is V2. The the output impedance is given by Z = R1 * R2 * (V1  V2)/(R1 * V2  R2 * (V1V2)) (I think I have the maths right. It is based on modelling the amplifier output as an ideal voltage source V0 in series with an impedance Z. It is then just a simple calculation of voltage dividers.) In your case, as I said, taking R1= 64 ohms might be a good choice. For R2, I would suggest something of the order of 5 or 10 times R1, so, say, about 600 ohms. Of course, you will probably want to choose R1 and R2 to be actual resistance values you have or can easily obtain. The important thing is to know exactly what they are. Much more typically, one would be measuring for an amplifier typically driving into an 8 ohm load, and so one would probably want to take R1 to be 8 ohms, and R2 maybe 47 ohms or so. The results will not be perfect, but I doubt that you really need to know the output impedance very accurately in any case. There is another method for measuring output impedance, involving the use of a second amplifier to driver a signal into the output of the amplifier under measurement. This is somewhat more involved to set up, even in the best of circumstances, and I don't think it is worth doing for your purposes. And in your case, it would be especially complicated to do it that way, because your amplifier is a circlotron, which means that neither of its output terminals is connected to ground. You would need to take special precautions to arrange to have a fully floating second amplifier to drive into the one you want to measure. It's just not worth it. The method with the R1 and R2 resistors I described above is the way to go, I think. Chris 
31st January 2013, 02:44 AM  #4 
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Join Date: Aug 2009
Location: College Station, TX

My reply #3 and couterculture's reply #2 came out more or less simultaneously. Our methods, and formulae are the same, but our notations are slightly different. His R2 is the same as my 1/(1/R1 + 1/R2).
Chris 
31st January 2013, 03:05 AM  #5  
diyAudio Member
Join Date: Jan 2009

re wire 901's
Quote:


31st January 2013, 05:49 AM  #6 
diyAudio Member
Join Date: Jan 2013

my first attemt to reply dissapeared, so here goes again, thanks counterculture and cn pope for the formulae...I will set up the apropriate tests. And about the bose speakers to J Michael Droke, the newer series boses are one ohms and have 3 mounting tabs, the older ones, (mine are series two), read out resistancewise at nearly eight ohms and have 4 mounting tabs. Looking into the box, there is a complex of wires, no series hookup. I will be drawing up a schematic of the exact wiring, I think theyr'e series/parrelled with the front speaker tied to the center of the arrangement, with a resistor across it's leads, because when I unhooked the speaker, the other speakers played a little weaker, and got louder when I shorted out the leads to the front.

31st January 2013, 11:51 AM  #7 
diyAudio Member
Join Date: May 2007

I told him several times in the other thread to maximise power, not voltage or current.

31st January 2013, 12:51 PM  #8 
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Join Date: Feb 2003
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the 2nd amp, power R tugging on the amp output, measuring the small V movement is the standard way really shoul;d doo a frequency sweep while you're at it
even with bridged/floating outputs it can be done with isolated test amp power supply trying to match power is usually not reasonable with very high feedback amps  the output Z will be milliOhms so the approximate current source load condition is the practical approach soundcards easily measure mVrms  add a op amp V gain stage if you need to 
31st January 2013, 12:54 PM  #9 
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Join Date: May 2007

Is it a high feedback amp?

31st January 2013, 01:09 PM  #10 
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Join Date: Feb 2003
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OK, I ws too quick with the SS based numbers  milliOhm is unlikely with tube amps
most tube amps try to achieve some damping ratio with nominal 8 Ohm speakers  the description in the 1st post is confusing so I can't really say, though followers are "high feedback" for tube circuits the point I'm tyring to make is that matching is unnecessary, in fact not often tried we can measure with even 100:1 mismatch between amp and test source Z with adequate resolution  with even cheap motherboard sound chipsets you typically get over 80 dB dynamic range Last edited by jcx; 31st January 2013 at 01:12 PM. 
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